MHB Can you find the two numbers?

[mathdad]

The sum of two numbers is 13 and their product is 40. Find the numbers.

Is this the correct set up?

Let x and y be the two numbers.

x + y = 13
xy = 40

 

[MarkFL]

The sum of two numbers is 13 and their product is 40. Find the numbers.

Is this the correct set up?

Let x and y be the two numbers.

x + y = 13
xy = 40

Yes, that's correct. (Yes)

 

[mathdad]

Yes, that's correct. (Yes)

I am going to use the MHB mainly for word problems. I want to increase my skills with word problems, particularly setting up the proper equations needed to find the answer(s).

 

[Greg]

Alternatively, factor 40

$$40=2^3\cdot5$$

and continue from there. Do you see how to continue?

 

[HallsofIvy]

Alternatively, factor 40

$$40=2^3\cdot5$$

and continue from there. Do you see how to continue?

Why? There was no requirement in the original problem that x and y be integers.

 

[mathdad]

Alternatively, factor 40

$$40=2^3\cdot5$$

and continue from there. Do you see how to continue?

Factoring 40 as 2^3 • 5 leads to the solution 8 and 5.

Yes?

 

[mathdad]

x + y = 13...A
xy = 40...B

xy = 40

x = 40/y

Plug into A.

x + y = 13

(40/y) + y = 13

40 + y^2 = 13y

y^2 -13y + 40 = 0

(y - 8)(y - 5) = 0

y = 8, y = 5

Plug into A or B to find x.

x + y = 13

x + 8 = 13

x = 13 - 8

x = 5

x + 5 = 13

x = 13 - 5

x = 8

Solution:

(5, 8) or (8, 5)

Correct?

 

[Greg]

Factoring 40 as 2^3 • 5 leads to the solution 8 and 5.

Yes?

Yes.

 

[MarkFL]

x + y = 13...A
xy = 40...B

xy = 40

x = 40/y

Plug into A.

x + y = 13

(40/y) + y = 13

40 + y^2 = 13y

y^2 -13y + 40 = 0

(y - 8)(y - 5) = 0

y = 8, y = 5

Plug into A or B to find x.

x + y = 13

x + 8 = 13

x = 13 - 8

x = 5

x + 5 = 13

x = 13 - 5

x = 8

Solution:

(5, 8) or (8, 5)

Correct?

While we know neither $x$ nor $y$ can be zero since their product is non-zero, I would in general tend to favor a substitution that doesn't involve division.

I would write:

$$y=13-x$$

And the substitute for $y$ as follows:

$$x(13-x)=40$$

$$13x-x^2=40$$

$$x^2-13x+40=0$$

Interestingly, factoring this quadratic requires us to essentially solve the original problem, that is to find two numbers whose sum is -13, and whose product is 40. So, let's use the quadratic formula instead:

$$x=\frac{13\pm\sqrt{13^2-4(1)(40}}{2(1)}=\frac{13\pm3}{2}\implies x\in\{5,8\}$$

From this we then conclude:

$$(x,y)\in\{(5,8),\,(8,5)\}$$

 

[mathdad]

While we know neither $x$ nor $y$ can be zero since their product is non-zero, I would in general tend to favor a substitution that doesn't involve division.

I would write:

$$y=13-x$$

And the substitute for $y$ as follows:

$$x(13-x)=40$$

$$13x-x^2=40$$

$$x^2-13x+40=0$$

Interestingly, factoring this quadratic requires us to essentially solve the original problem, that is to find two numbers whose sum is -13, and whose product is 40. So, let's use the quadratic formula instead:

$$x=\frac{13\pm\sqrt{13^2-4(1)(40}}{2(1)}=\frac{13\pm3}{2}\implies x\in\{5,8\}$$

From this we then conclude:

$$(x,y)\in\{(5,8),\,(8,5)\}$$

Interesting second method.

 

[Greg]

Why? There was no requirement in the original problem that x and y be integers.

13 and 40 are integers so it makes sense to investigate the possibility that x and y are integers. The problem is quite simple due to the size of the numbers involved; if the numbers were larger it would possibly make more sense to use the method demonstrated by Mark. If they were not integers then Mark's method of solution would definitely be preferred.

 

[mathdad]

I thank everyone for your help and suggestions.